USTET Mathematics — Calculus — Limits, Derivatives & IntegralsFlash Cards

Evaluate: lim(x→3) (2x + 5)

Step 1: Apply the limit of a linear function theorem: lim(x→a) (mx + b) = ma + b Step 2: Substitute a = 3, m = 2, b = 5 Step 3: lim(x→3) (2x + 5) = 2(3) + 5 = 6 + 5 = 11 Answer: 11

Find the derivative of f(x) = x⁴ using the power rule

Step 1: Apply the power rule: d/dx(xⁿ) = nxⁿ⁻¹ Step 2: For f(x) = x⁴, we have n = 4 Step 3: f'(x) = 4x⁴⁻¹ = 4x³ Answer: f'(x) = 4x³

Evaluate: ∫ 3x² dx

Step 1: Apply the power rule for integration: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C Step 2: For 3x², factor out the constant: ∫ 3x² dx = 3∫ x² dx Step 3: ∫ x² dx = x²⁺¹/(2+1) + C = x³/3 + C Step 4: 3∫ x² dx = 3(x³/3) + C = x³ + C Answer: x³ + C

Solve: lim(x→2) (x² - 4)/(x - 2)

Step 1: Direct substitution gives 0/0 (indeterminate form) Step 2: Factor the numerator: x² - 4 = (x + 2)(x - 2) Step 3: Simplify: (x² - 4)/(x - 2) = [(x + 2)(x - 2)]/(x - 2) = x + 2 (for x ≠ 2) Step 4: Now evaluate: lim(x→2) (x + 2) = 2 + 2 = 4 Answer: 4

Find the derivative of h(x) = 5x³ - 2x + 7

Step 1: Apply the sum/difference rule: derivative of sum = sum of derivatives Step 2: d/dx(5x³) = 5 · 3x² = 15x² (constant multiple and power rule) Step 3: d/dx(-2x) = -2 · 1 = -2 (constant multiple rule) Step 4: d/dx(7) = 0 (derivative of constant) Step 5: h'(x) = 15x² - 2 + 0 = 15x² - 2 Answer: h'(x) = 15x² - 2

When do you use the quotient rule for derivatives?

Use the quotient rule when finding the derivative of f(x)/g(x) where both f(x) and g(x) are functions. Formula: If h(x) = f(x)/g(x), then h'(x) = [g(x)f'(x) - f(x)g'(x)]/[g(x)]² Example: For h(x) = x²/(x+1) f(x) = x², f'(x) = 2x g(x) = x+1, g'(x) = 1 h'(x) = [(x+1)(2x) - x²(1)]/(x+1)² = (2x² + 2x - x²)/(x+1)² = (x² + 2x)/(x+1)²

Evaluate: ∫₁³ 2x dx using the Fundamental Theorem of Calculus

Step 1: Find the antiderivative F(x): ∫ 2x dx = 2 · x²/2 + C = x² + C Step 2: Apply FTC Part 2: ∫ₐᵇ f(x)dx = F(b) - F(a) Step 3: ∫₁³ 2x dx = [x²]₁³ = F(3) - F(1) Step 4: F(3) = 3² = 9, F(1) = 1² = 1 Step 5: F(3) - F(1) = 9 - 1 = 8 Answer: 8

Find the derivative of f(x) = (2x + 1)(x² - 3) using the product rule

Step 1: Product rule: If h(x) = f(x)g(x), then h'(x) = f(x)g'(x) + f'(x)g(x) Step 2: Let f(x) = 2x + 1, g(x) = x² - 3 Step 3: Find derivatives: f'(x) = 2, g'(x) = 2x Step 4: Apply product rule: h'(x) = (2x + 1)(2x) + (2)(x² - 3) Step 5: Simplify: h'(x) = 4x² + 2x + 2x² - 6 = 6x² + 2x - 6 Answer: h'(x) = 6x² + 2x - 6

Evaluate: lim(x→0) (sin x)/x (without using L'Hôpital's rule)

This is a standard limit that equals 1. Step 1: This is a fundamental trigonometric limit Step 2: lim(x→0) (sin x)/x = 1 Note: This limit is proven using the squeeze theorem and geometric arguments. It's a standard result used to derive other trigonometric limits. Answer: 1

Find ∫ (3x² - 4x + 5) dx

Step 1: Integrate term by term using sum/difference rule Step 2: ∫ 3x² dx = 3 · x³/3 + C₁ = x³ + C₁ Step 3: ∫ -4x dx = -4 · x²/2 + C₂ = -2x² + C₂ Step 4: ∫ 5 dx = 5x + C₃ Step 5: Combine: ∫ (3x² - 4x + 5) dx = x³ - 2x² + 5x + C Answer: x³ - 2x² + 5x + C

Why does the derivative of a constant equal zero?

Step 1: A constant function f(x) = c has the same value for all x Step 2: The derivative measures the rate of change Step 3: Since a constant doesn't change, its rate of change is zero Step 4: Using the definition: f'(x) = lim(h→0) [f(x+h) - f(x)]/h = lim(h→0) [c - c]/h = lim(h→0) 0/h = 0 Example: If f(x) = 7, then f'(x) = 0 because 7 never changes Answer: Because constants have zero rate of change

Solve: lim(x→∞) (2x + 3)/(x - 1)

Step 1: For limits at infinity with rational functions, divide by highest power of x Step 2: Divide numerator and denominator by x: lim(x→∞) (2x + 3)/(x - 1) = lim(x→∞) (2 + 3/x)/(1 - 1/x) Step 3: As x→∞, 3/x→0 and 1/x→0 Step 4: lim(x→∞) (2 + 3/x)/(1 - 1/x) = (2 + 0)/(1 - 0) = 2/1 = 2 Answer: 2

Find the derivative of y = x⁻³ + 2x⁻¹

Step 1: Rewrite using negative exponents: y = x⁻³ + 2x⁻¹ Step 2: Apply power rule: d/dx(xⁿ) = nxⁿ⁻¹ Step 3: d/dx(x⁻³) = -3x⁻³⁻¹ = -3x⁻⁴ Step 4: d/dx(2x⁻¹) = 2(-1)x⁻¹⁻¹ = -2x⁻² Step 5: y' = -3x⁻⁴ - 2x⁻² = -3/x⁴ - 2/x² Answer: y' = -3x⁻⁴ - 2x⁻² or -3/x⁴ - 2/x²

Evaluate: ∫₀² (x² + 1) dx

Step 1: Find antiderivative: ∫ (x² + 1) dx = x³/3 + x + C Step 2: Apply FTC: ∫₀² (x² + 1) dx = [x³/3 + x]₀² Step 3: Evaluate at upper limit x = 2: 2³/3 + 2 = 8/3 + 2 = 8/3 + 6/3 = 14/3 Step 4: Evaluate at lower limit x = 0: 0³/3 + 0 = 0 Step 5: Subtract: 14/3 - 0 = 14/3 Answer: 14/3

What's the relationship between derivatives and integrals?

They are inverse operations (Fundamental Theorem of Calculus): Step 1: If F(x) is the antiderivative of f(x), then F'(x) = f(x) Step 2: If f(x) is the derivative of F(x), then ∫ f(x) dx = F(x) + C Example: f(x) = 2x, F(x) = x² - Derivative: d/dx(x²) = 2x ✓ - Integral: ∫ 2x dx = x² + C ✓ They "undo" each other like addition and subtraction Answer: Derivatives and integrals are inverse operations

Find the derivative of f(x) = √x using the power rule

Step 1: Rewrite using fractional exponent: f(x) = √x = x^(1/2) Step 2: Apply power rule: d/dx(xⁿ) = nxⁿ⁻¹ Step 3: f'(x) = (1/2)x^(1/2-1) = (1/2)x^(-1/2) Step 4: Rewrite in radical form: f'(x) = (1/2)x^(-1/2) = 1/(2√x) Step 5: Simplify: f'(x) = 1/(2√x) Answer: f'(x) = 1/(2√x) or (1/2)x^(-1/2)

Evaluate: lim(x→4) (√x - 2)/(x - 4)

Step 1: Direct substitution gives 0/0 (indeterminate) Step 2: Multiply by conjugate: (√x - 2)/(x - 4) · (√x + 2)/(√x + 2) Step 3: Numerator: (√x - 2)(√x + 2) = x - 4 Step 4: Simplify: (x - 4)/[(x - 4)(√x + 2)] = 1/(√x + 2) for x ≠ 4 Step 5: Evaluate: lim(x→4) 1/(√x + 2) = 1/(√4 + 2) = 1/(2 + 2) = 1/4 Answer: 1/4

Find ∫ 1/x² dx

Step 1: Rewrite using negative exponent: 1/x² = x⁻² Step 2: Apply power rule: ∫ xⁿ dx = x^(n+1)/(n+1) + C where n ≠ -1 Step 3: ∫ x⁻² dx = x^(-2+1)/(-2+1) + C = x⁻¹/(-1) + C Step 4: Simplify: -x⁻¹ + C = -1/x + C Answer: -1/x + C

Apply the product rule to find d/dx[(x² + 1)(2x - 3)]

Step 1: Product rule: d/dx[f(x)g(x)] = f(x)g'(x) + f'(x)g(x) Step 2: Let f(x) = x² + 1, g(x) = 2x - 3 Step 3: Find derivatives: f'(x) = 2x, g'(x) = 2 Step 4: Apply formula: d/dx[(x² + 1)(2x - 3)] = (x² + 1)(2) + (2x)(2x - 3) Step 5: Expand: = 2x² + 2 + 4x² - 6x = 6x² - 6x + 2 Answer: 6x² - 6x + 2

Common mistake: Why can't you just integrate each part separately in ∫ x·eˣ dx?

Step 1: This is a PRODUCT of functions: x and eˣ Step 2: ∫ [f(x)·g(x)] dx ≠ ∫ f(x) dx · ∫ g(x) dx Step 3: Wrong approach: ∫ x dx · ∫ eˣ dx = (x²/2)·(eˣ) ❌ Step 4: Correct approach: Use integration by parts Step 5: ∫ u dv = uv - ∫ v du where u = x, dv = eˣ dx Remember: Integration distributes over ADDITION, not MULTIPLICATION Answer: Because integration doesn't distribute over products - need integration by parts